các bạn giúp mình gpt lượng giác nhé

T

truongduong9083

Chào bạn

Câu d
Biến đổi thành:
$$(sin^2x+cos^2x)(sin^3x+cos^3x) = 2(sin^5x+cos^5)$$
$$\Leftrightarrow sin^5x+cos^5x = sin^3xcos^2x+cos^3xsin^2x$$
$$\Leftrightarrow sin^3x(sin^2x-cos^2x)-cos^3x(sin^2x-cos^2x) = 0$$
$$\Leftrightarrow cos2x(cos^3x-sin^3x) = 0$$
Đến đây cơ bản rồi nhé
 
N

newstarinsky

$d) sin^3x-2sin^5x=2cos^5x-cos^3x\\
\Leftrightarrow sin^3x(1-2sin^2x)=cos^3x(2cos^2x-1)\\
\Leftrightarrow cos2x(cos^3x-sin^3x)=0\\
\Leftrightarrow cos2x(cosx-sinx)(1+sinx.cosx)=0$

$b)sin^3x-sinx+cos^3x+cosx=0\\
\Leftrightarrow -cos^2x.sinx+cos^3x+cosx=0\\
\Leftrightarrow cosx(1-sinx.cosx+cosx^2x)=0\\
\Leftrightarrow cosx(3+cos2x-sin2x)=0$

$a) (1-sin^2x)^2-1+2sin^2x+2sin^6x=0\\
\Leftrightarrow 2sin^6x+sin^4x=0\\
\Leftrightarrow sin^4x(2sin^2x+1)=0$

$e) \dfrac{2sinx}{cosx}+\dfrac{cos2x}{sin2x}=sin2x+
\dfrac{1}{sin2x}\\
\Leftrightarrow 4sin^2x+cos2x=sin^22x+1\\
\Leftrightarrow 2-2cos2x+cos2x=2-cos^22x\\
\Leftrightarrow cos^22x-cos2x=0$
 
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S

sky_fly_s2

câu c nhé!!!!

c.
$\Leftrightarrow (1-cos2x)/2=cos^22x+3(cos6x+1)/2$
$\Leftrightarrow 3cos6x+2cos^22x+cos2x+2=0$
$\Leftrightarrow 4cos^32x+2cos^22x-2cos2x+2=0$
 
C

cafekd


$2cos^2(\frac{\pi}{4}-3x)-4cos4x-15sin2x=21$

\Leftrightarrow $(sin3x + cos3x)^2 - 4(1-2sin^22x) - 15sin2x - 21 = 0$

\Leftrightarrow $1 + sin6x - 4 + 8sin^22x - 15sin2x - 21 = 0$

\Leftrightarrow $3sin2x - 4sin^32x + 8sin^22x - 12sin2x - 24 = 0$

\Leftrightarrow $sin2x = -1 $

\Leftrightarrow $2x = \dfrac{-\pi}{2} + k2\pi$

\Leftrightarrow $x = \dfrac{-\pi}{4} + k\dfrac{\pi}{2}$




 
H

haohaohao_tq

Bài này nữa nhé!

$\int \dfrac{e^{2x}dx}{e^x + 6e^{-x} - 5}$
 
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